PART 10 Summary of organic reaction mechanisms - A mechanistic introduction to organic chemistry and explanations of different types of organic reactions

Doc Brown's Chemistry Advanced Level Pre-University Chemistry Revision Study Notes for UK KS5 A/AS GCE advanced level organic chemistry students US K12 grade 11 grade 12 organic chemistry

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10.3 Reaction mechanisms of ALKENES - Electrophilic addition of hydrogen bromide

Part 10.3 ALKENES - introduction to the reaction mechanisms of alkenes.

Electrophilic addition of hydrogen bromide [HBr(conc. aq) and HBr(g/non-polar solvent)] to form halogenoalkanes.

These revision notes include full diagrams and explanation of the ionic electrophilic addition reaction mechanisms of alkenes and the 'molecular' equation and reaction conditions and other con-current reaction pathways and products are also explained.

Sub-index for this page

10.3.1 Introduction - the bonding in alkenes and reactivity

10.3.2 The addition of hydrogen bromide to an alkene

10.3.3 The electrophilic addition mechanism of anhydrous hydrogen bromide to an alkene

10.3.4 The electrophilic addition mechanism of aqueous hydrogen bromide (hydrobromic acid) to alkene

10.3.5 Further comments on the electrophilic addition of hydrogen bromide to alkene e.g. evidence of ionic mechanism, carbocation stability and Markovnikov's rule on orientation of products

See also

Bonding in alkenes, alkene reactivity compared to alkanes, electrophilic addition reaction with hydrogen halides

• The reactive nature of alkenes

• Alkenes are reactive molecules, particularly when compared to alkanes.

• They are reactive towards electron pair accepting electrophiles because of the high density of negative electron charge associated with the pi electrons of the double bond (diagram below)

• The π and σ bonds of the C=C double bond are shown in the diagram

• The sigma bond is present between all the atoms in organic molecules. However in alkenes, the double bond consists of a sigma bond (σ) and a pi bond (π). Two electrons are in the molecular orbital of the sigma bond which is directed linearly between the two carbon atoms.

• BUT, the other two electrons of the C=C double bond are in the two pi orbitals which lie above and below the plane of the sigma bond of the >C-C< system (1 electron per pi electron cloud).

• It is the electrons of these pi orbitals that are susceptible to electrophilic attack by an electron pair acceptor - an electrophile.

• However alkenes can also readily undergo free radical reactions  e.g. their peroxide catalysed polymerisation to form a poly(alkene) and these reactions also involve the interaction of free radicals with the π pi electrons.

• The electrophilic addition reactions of alkenes are compared with the nucleophilic addition to carbonyl compounds in the aldehydes and ketones section.

• Examples of the addition of hydrogen bromide to alkenes

  • (i) ethene + hydrogen bromide ==> bromoethane

  • + HBr ===>  

  • (ii) propene + hydrogen bromide ==> 2-bromopropane

  • + HBr ===>

    • some of the isomer 1-bromopropane is also formed in this reaction,

    • the minority product is from a con-current reaction.

• What is the reaction mechanism for the addition of hydrogen bromide to an alkene?

• Does the mechanism change if the solvent is changed?

• Do the products of the reaction depend on the solvent used?

• Can isomeric products be formed in the addition of hydrogen bromide to an alkene?

10.3.3 The electrophilic addition of anhydrous hydrogen bromide to an alkene

• NON-AQUEOUS media: [mechanism diagram 3 below]

  • Addition will also occur if the alkene is mixed with hydrogen bromide gas, or the HBr gas is dissolved in a non-polar organic solvent and mixed with the alkene.

  • The addition of pure hydrogen bromide gas is a simpler situation compared to the addition of aqueous hydrobromic acid.

mechanism 3 - electrophilic addition of hydrogen bromide to an alkene in non-aqueous media

• In this case, for step (1), the attacking electrophile is the already polarised hydrogen bromide molecule, H^δ+Br^δ-, which splits heterolytically to protonate the alkene, forming the carbocation and a bromide ion. The HBr molecule is an electrophile because it accepts a pair of electrons from the alkene π bond to form the new C-H bond.

• In step (2) the bromide ion formed in step (1) rapidly combines with the carbocation to form the bromoalkane. The bromide ion donates a pair of electrons to form the new C-Br bond.

10.3.4 Electrophilic addition of aqueous hydrogen bromide (hydrobromic acid) to alkene

• AQUEOUS media: [see mechanism 39 and 39a below]

  • If a liquid alkene is mixed with, or a gaseous alkene is bubbled through, concentrated hydrobromic acid, HBr_(aq) (hydrogen bromide dissolved in water) a bromoalkane is formed. 

  • overall reaction: R₂C=CR₂  +  HBr ===> R₂CH-CBrR₂ 

  • In water, hydrogen bromide is a strong acid i.e. completely ionises to give the oxonium ion and bromide ion.

    • HBr_(g)  +  H₂O_(l)  ===>  H₃O⁺_(aq)  +  Br^-_(aq) 

    • The aqueous media complicates the reaction compared to the addition of pure hydrogen bromide gas.

mechanism diagram 39 - electrophilic addition of hydrogen bromide to an alkene in aqueous media

• In the acid solution via step (1) the H₃O⁺ or oxonium ion (hydrated proton) is the 'attacking electrophile' and protonates the alkene to form the intermediate positive carbocation R₂CHCR₂⁺. The oxonium ion is an electrophile because it accepts a pair of electrons from the alkene π bond to form the new C-H bond.

• In step (2) the (already present) negative bromide ion rapidly combines with the carbocation to form the bromoalkane product. The bromide ion donates a pair of electrons to form the new C-Br bond.

• With the high concentration of water present, a water molecule could also interact with the carbocation to eventually form a small amount of the alcohol R₂CHCR₂OH, this again provides evidence of an ionic mechanism with a carbocation intermediate.

• Step2 in mechanism diagram 39a: R₂CHCR₂⁺ + H₂O  === > R₂CHCR₂OH  + H⁺

10.3.5 Further comments on the electrophilic addition of hydrogen bromide to alkene

• FURTHER COMMENTS

  • EVIDENCE for an IONIC MECHANISM

    • Below is a general comment for all the electrophilic addition reactions of alkenes.

    • If the reaction is carried out in the presence of other negative ions e.g. chloride ion from adding sodium chloride salt to an aqueous reaction mixture, then some chloroalkane is produced via step (2).

      • R₂CH-CR₂⁺  +  Cl^-  ===>  R₂CH-CR₂Cl 

        • Without a carbocation intermediate formed it is difficult to explain the formation of such products.

        • In fact any anion present, X^-, produces some R₂CH-CR₂X

  • A symmetrical alkene is when the atoms/groups are the same on either side of the C=C double bond.

    • e.g. ethene H₂C=CH₂ or but-2-ene CH₃-CH=CH-CH₃ 

    • This means which ever way round the HX addition takes place onto the double bond, you always get the same product.

  • An non-symmetrical alkene is when the atoms/groups are NOT the same on either side of the C=C double bond e.g.

    • propene CH₃-CH=CH₂, methylpropene (CH₃)₂C=CH₂ or but-1-ene CH₂=CH-CH₂-CH₃ 

    • This means that when addition to the double bond with a non-symmetrical reagent itself, e.g. like H-X, you have the possibility of two different isomeric addition products.

      • e.g. CH₃-CH=CH₂  +  H-X  ===>  CH₃-CHX-CH₃  or  CH₃-CH₂-CH₂-X

      • Which begs the questions, which isomer predominates? and why?

  • The Markownikoff (Markovnikov) rule predicts which isomer is likely to predominate for adding a non-symmetrical reagent to a non-symmetrical alkene and the rule can be stated in various ways but the IUPAC definition of 1997 states:

    • For the heterolytic addition of a polar molecule to an alkene (or alkyne), the more electronegative (the more nucleophilic like OH^-, H₂O or Br^- etc.) atom (or part) of the polar molecule becomes attached to the carbon atom bearing the smaller number of hydrogen atoms.

      • This is usually step 2 in the electrophilic addition mechanisms

    • Or, you can say the least electronegative (more electrophilic like Br⁺ or H⁺ etc.) will attach to the carbon atom bonded with the most H atoms.

      • This is usually step 1 in the electrophilic addition mechanisms, noting which carbocation is the most stable.

    • BUT the 'rule' only applies to the ionic electrophilic addition mechanism, you can get the opposite effect in free radical addition in the presence of peroxides!

  • The orientation of the products from non-symmetrical addition (HX or Br_2(aq) see later) is governed by the stability of the carbocation intermediate formed by the protonation of the alkene by the attacking H-X electrophile, and explains the Markownikoff rule.

    • The order of carbocation stability is tertiary > secondary > primary, because alkyl groups give a slight electron donating inductive effect (+I) via the attraction of the positively charged carbon atom.

    • This spreads the positive charge of the carbocation and gives the carbocation more stability by lowering its potential energy.

    • It is a general rule of physics that spreading out electric charge lowers the potential energy and increases the stability of a situation.

    • This 'inductive' electron cloud shift and relative stability is illustrated in the mechanism diagrams 55a/55 below.

• The most stable carbocation will be the one most likely to exist with a sufficient life-time to be hit by the electron pair donating ion (e.g. X^-) or any other electron pair donor, including water (see addition of bromine water).

• NOTE: The positive carbon of the most stable carbocation, has attached to it, the most alkyl groups and the least hydrogen atoms.

  • e.g. from protonating propene CH₃CH=CH₂ you expect the carbocation stability to be ...

    •  CH₃CH⁺CH₃ (sec) > CH₃CH₂CH₂⁺ (prim)

  • or from protonating 2-methylbut-2-ene (CH₃)₂C=CHCH₃ you expect the carbocation stability to be ...

    • (CH₃)₂C⁺CH₂CH₃ (tert) > (CH₃)₂CHC⁺HCH₃ (sec)

• So for adding HX to a non-symmetrical alkene you would expect the major isomer to be e.g.

  • from propene, CH₃CH=CH₂ you expect mainly CH₃CHX-CH₃

    • and some CH₃CH₂-CH₂X

  • from methylpropene, (CH₃)₂C=CH₂ you expect mainly (CH₃)₂CX-CH₃  

    • and some (CH₃)₂CH-CH₂X

  • from 2-methylbut-2-ene you expect mainly (CH₃)₂CXCH₂CH₃ > and some (CH₃)₂CHCHXCH₃

  • from but-1-ene, CH₂=CHCH₂CH₃ you expect mainly CH₃-CHXCH₂CH₃  

    • with some XCH₂-CH₂CH₂CH₃ 

• What happens in terms of optical isomers/activity if the product has a chiral carbon^*?

  • An example of a  chiral carbon results from when four different atoms or groups (a to d) is bonded to the same carbocation i.e. ^*Cabcd

  • So the carbocation has a plane of symmetry due to the trigonal planar arrangement of the bonds around the carbon atom carrying the positive charge.

  • This symmetrical arrangement means that if the product is potentially optically active, a racemic mixture will be formed because the e.g. bromide ion, can add with equal probability on both sides of the carbocation.

  • This will result in equal quantities of the optical isomers (enantiomers), giving an optically inactive racemic mixture.

  • e.g. but-1-ene, CH₂=CHCH₂CH₃ on adding HX, will give a racemic mixture of the optical isomers of CH₃^*CHXCH₂CH₃ with some XCH₂CH₂CH₂CH₃ which is incapable of optical isomerism because it does not have a chiral carbon.

• Extra comment !

  • Free radical addition of hydrogen bromide

    • The addition of hydrogen bromide using a peroxide catalyst produces 'anti-Markownikoff' addition.

      • e.g. propene, CH₃CH=CH₂ produces mainly CH₃CH₂-CH₂Br

      • but only hydrogen bromide gives this peroxide effect, the addition of H-Cl, Br_2(aq) and H₂SO₄ etc. all broadly follow the Markownikoff addition rule.

  • Addition of mixed halogen compounds (inter-halogen compounds), such as iodine(I) chloride ICl, will also add to the alkene double bond. 

    • e.g. CH₃CH=CH₂  +  ICl  ===>  CH₃CHI-CH₂Cl  or  CH₃CHCl-CH₂I 

    • From the Markownikoff rule 2-chloro-1-iodopropane should be the principal product because chlorine is more electronegative than iodine, so think of it as the addition of I^δ+-Cl^δ-.